Using a weight function for inner product/normalization

1. Nov 28, 2004

kakarukeys

if we use weight function for inner product/normalization of wavefunctions, what will the momentum operator look like?

$$W(q,t)dq = dx$$

$$\int \psi'^*(q,t)\psi'(q,t) W(q,t)dq = \int \psi^*(x,t)\psi(x,t) dx$$

2. Dec 2, 2004

jujio77

I may be wrong, but doesn't the weight function have something to do with the coordinate system you are in. If we look at a problem that has spherical symmetry our weight function is
$W=r^2sin\theta$.

We know the momentum operator in this case.

3. Dec 3, 2004

dextercioby

In a separable Hilbert space the choise of basis for the space is not unique.In your case of Hilbert space (the space of square modulus integrable functions) you can pick either of the 2 representations u like:momentum/coordinate.Let's say u have the more common coordinate representation.The simplest form for the Hamilton operator is in the rectangular coordinates,yet some systems may have coordinate symmetries,e.g.spherical,cylindrical,hyperboloidal,...The change of coordinates in the same representation automatically alters the other representation's operators and variables.
If u change the "x","p_x" will change as well,but the fundamental commutation relation is left invariant.

To conclude,my guess is that weighing functions are the Jacobian of the (usually orthogonal) coordinate transformation within the same representation,and their connection wrt to mometum operators is not immediate in the coordinate representation but is relevant in the momentum representation.

Last edited: Dec 3, 2004
4. Dec 10, 2004

kakarukeys

Thanks, by the time there were replies to my thread.
I already figured it out myself.
Yes weight function has something to do with coordinate transformation.

I only had one question not figured out, but this is mathematical
Given two equivalent descriptions of the physics related by a coordinate transformation, one of them has a weight function always > 0 in its inner product.

$$\int \psi'^*_A(q,t)\psi'_B(q,t) W(q,t)dq = \int \psi^*_A(q,t)\psi_B(q,t) dq$$
upper/lower limits are the same.

Is $$\psi'(q,t) = \frac{C}{\sqrt{W(q,t)}}\psi(q,t)$$ the only possible transformation? $$(|C|^2 = 1)$$

5. Dec 10, 2004

seratend

C=exp(-if(q,t)) also works where f is any "good" function.

Seratend.

6. Dec 11, 2004

kakarukeys

I meant besides this kind of manipulation